There are constants $\alpha$ and $\beta$ such that $\frac{x-\alpha}{x+\beta} = \frac{x^2-80x+1551}{x^2+57x-2970}$. What is $\alpha+\beta$?
Solution: The numerator $x^2 - 80x + 1551$ factors as $(x - 47)(x - 33)$, and the denominator $x^2 + 57x - 2970$ factors as $(x - 33)(x + 90)$, so \[\frac{x^2 - 80x + 1551}{x^2 + 57x - 2970} = \frac{(x - 47)(x - 33)}{(x - 33)(x + 90)} = \frac{x - 47}{x + 90}.\]Then $\alpha = 47$ and $\beta = 90$, so $\alpha + \beta = 47 + 90 = \boxed{137}$.

We can also solve the problem using Vieta's formulas, which states that the sum of the roots of the quadratic $ax^2 + bx + c = 0$ is $-b/a$.  The only way that the right-hand side $\frac{x^2-80x+1551}{x^2+57x-2970}$ can simplify to the left-hand side $\frac{x-\alpha}{x+\beta}$ is if $x^2-80x+1551$ and $x^2+57x-2970$ have a root in common.  Call this common root $\gamma$.

Then the roots of $x^2 - 80x + 1551 = 0$ are $\alpha$ and $\gamma$, so $\alpha + \gamma = 80$.  Similarly, the roots of $x^2 + 57x - 2970 = 0$ are $-\beta$ and $\gamma$, so $-\beta + \gamma = -57$.  Subtracting these equations, we get $\alpha + \beta = 80 - (-57) = \boxed{137}$.